A charged capacitor $C_1$ is discharged through a resistance $R$ by putting switch S in position $1$ of circuit shown in figure. When discharge current reduces to $I_0$ the switch is suddenly shifted to position $2$. Calculate the amount of heat liberated in resistance $R$ starting from this instant. 

Capacitor $C_1$ will discharge according to the equation, 

$q=q_0{e}^{-t/{\tau }_C}$ …(i)

Here, ${\tau }_C=C_{1}R$

and discharging current:

$i=\frac{-dq}{dt}=\frac{q_0}{{\tau }_C}·e^{-t/{\tau }_C}=\frac{q_0{e}^{-t/{\tau }_C}}{C_{1}R}$ ….(ii)

At the given instant $i=i_0$

Therefore for equation (ii) 

$q_0{e}^{-t/{\tau }_C}=i_0{C}_{1}R$ at this instant 

or charge $C_1$ at this instant will be 

$q=\left(i_0{C}_{1}R\right)$ [From equation (ii)]

Now this current $q$ will later on distribute in $C_1$ and $C_2$.

$\therefore U_i=\frac{q^2}{2C_1}and\frac{q^2}{2\left(C_1+C_2\right)}=U_f$

Heat generated in resistance ,

$H=U_i-U_f$

Substituting values of $q:U_i\text{ and }{U}_f$ we get, 

$H=\frac{{\left(I_{0}R\right)}^2{C}_1{C}_2}{2\left(C_1+C_2\right)}$